“或”-typed对象的访问属性

Question

如何在下面的代码中访问'y‘- attriubte？

export type AutogeneratedResult = {
  a: string
} | { a: string, b: string } | { y: string };

function foo(): AutogeneratedResult {
  return {
    y: 'test'
  }
}


const x = foo();

console.log(x.y);  // <---- Syntax error

我不知道怎样才能解决这个问题。

打字稿-游乐场：https://www.typescriptlang.org/play?#code/KYDwDg9gTgLgBDAnmYcCCBXGEDmwB2wUAhjMACYBKwAzhgDbwC8cA3gFBxzEBccNMKAEt8OdgF84AHzbc+A4aIA0cAEbzBInHEkzWcRBsXbxAbnbsAZhnwBjGEIj44liBAAUASj6ZseQiRkVLQM8BxcUMAwGFDO4VwGfADkZAJJnDoSFuy2TgJwIHAsrh6e5jl5EPTAAHT0uO4gNYhlQA

Answer 1

要么您知道将要返回的内容，然后返回精确的类型：

游乐场

type ChoiceA = { a: string };
type ChoiceB = { a: string, b: string };
type ChoiceC = { y: string };

export type AutogeneratedResult = ChoiceA | ChoiceB | ChoiceC;

function foo(): ChoiceC {
  return {
    y: 'test'
  };
}

const x = foo();

console.log(x.y);

要么您必须检查返回类型：

游乐场

type ChoiceA = { a: string };
type ChoiceB = { a: string, b: string };
type ChoiceC = { y: string };

export type AutogeneratedResult = ChoiceA | ChoiceB | ChoiceC;

function foo(): AutogeneratedResult {
  return {
    y: 'test'
  };
}

const x = foo();

if ('y' in x) {
    console.log(x.y);
}